MATH 251.504Practice Problems for Examination 1Fall 20061. For which values of a are the vectors ha, 3, −9i and ha, −4a, −4i orthogonal?2. Determine whether or not the points P (3, 7, 1), Q(4, 2, 1), R(−1, 3, 1), and S(0, 0, 2) liein the same plane.3. Find symmetric equations for the line that is contained in both the plane z = x andthe tangent plane to the surface z = 2x2+ 5y2at the point (1, 1, 7).4. Give an example of three-dimensional vectors a, b, and c such that a ×b = a ×c butb 6= c.5. Determine whether the following statement is true or false: For every function f :R2→ R which is continuous at (0, 0) the function g(x, y) = (x − 1)(y − 1)f(x, y) iscontinuous at (1, 1).6. Show that the limit doesn’t exist in each of the following two cases.(a) lim(x,y)→(0,0)x(y + 3x)5x2+ y2(b) lim(x,y)→(1,0)x2− 2x − y2+ 1x2− 2x + y2+ 17. Find fxx, fxyand fyy.(a) f(x, y) = (1 − x2)(1 − 2x3)(b) f(x, y) = ln(x2+ e2y)(c) f(x, y) = ye−2x+ cos(y2)8. Find an equation for the tangent plane to the surface z = xyexat the point (1, 1, e).9. Find the partial derivatives∂w∂sand∂w∂twhere w = 5xyz + ln(x −y) and x = st, y = t2,and z = 4t.10. Give an example of a function f : R2→ R such that fx(0, 0) and fy(0, 0) exist but f isnot continuous at (0, 0).111. Use differentials to approximate the number√353√215.12. In which direction does the function f (x, y, z) = zey+ xy2decrease the fastest at thepoint (1, 1, 1)?13. Classify the quadric surfaces x2− y2− 2y − z2= 5 and y − 4 = z2− x2+ 4x.14. Find an equation for the normal line to the surface z = cos(y) sin(x) at the point(π2,π4, 0).Solutions1. The two vectors are orthogonal precisely when the dot product ha, 3, −9i·ha, −4a, −4i =a2− 12a + 36 = (a − 6)2is zero, that is, when a = 6.2. It is equivalent to determine whether or not the vectors−→P Q = h1, −5, 0i,−→P R =h−4, −4, 0i, and−→P S = h−3, −7, 1i are coplanar. They are not coplanar because thescalar triple product−→P Q · (−→P R ×−→P S) =1 −5 0−4 −4 0−3 −7 1= 1(−4) − (−5)(−4) + 0 = −24is not equal to zero.3. Set f (x, y) = 2x2+ 5y2. Then fx(x, y) = 4x and fy(x, y) = 10y, and in particularfx(1, 1) = 4 and fy(1, 1) = 10. Therefore the equation of the tangent plane at (1, 1, 7)isz − 7 = fx(1, 1)(x − 1) + fy(1, 1)(y − 1) = 4(x − 1) + 10(y − 1)or 4x + 10y − z = 7, and it has h4, 10, −1i as a normal vector. The plane z = xhas h−1, 0, 1i as a normal vector, and so the line of intersection of the two planes hasdirection vector h4, 10, −1i × h−1, 0, 1i = h10, −3, 10i. Setting x = z = 0 we see fromthe equation 4x + 10y −z = 7 that (0,710, 0) is a p oint on the line of intersection of thetwo planes. Thus symmetric equations for this line of intersection arex10=y −710−3=z10.24. Take for example a = b = i and c = −i.5. The statement is false. Consider for example the functionf(x, y) =1(x − 1)(y − 1)if (x, y) 6= (1, 1),0 if (x, y) = (1, 1).6. (a) Approaching (0, 0) along the x-axis we have the limit limx→03x25x2=35, while ap-proaching (0, 0) along the y-axis we have the limit limy→00y2= 0, and so the givenlimit doesn’t exist.(b) Approaching (1, 0) along the x-axis we have the limit limx→1x2−2x+1x2−2x+1= 1, whileapproaching (1, 0) along the line x = 1 we have the limit limy→0−y2y2= −1, and so thegiven limit doesn’t exist.7. (a) fxx(x, y) = −2 − 12x + 40x3, fxy(x, y) = 0, fyy(x, y) = 0.(b) fxx(x, y) =2x2+e2y−4x2(x2+e2y)2, fxy(x, y) =−4xe2y(x2+e2y)2, fyy(x, y) =4e2yx2+e2y−4e4y(x2+e2y)2.(c) fxx(x, y) = 4ye−2x, fxy(x, y) = −2e−2x, fyy(x, y) = −2 sin(y2) − 4y2cos(y2).8. Set f(x, y) = xyex. Then fx(x, y) = (x + 1)yexand fy(x, y) = xyex, and in particularfx(1, 1) = 2e and fy(1, 1) = e. Therefore the equation of the tangent plane at (1, 1, e)isz − e = fx(1, 1)(x − 1) + fy(1, 1)(y − 1) = 2e(x − 1) + e(y − 1)or 2ex + ey − z = 2e.9. We have∂w∂s=∂w∂x∂x∂s+∂w∂y∂y∂s+∂w∂z∂z∂s=5yz +1x − yt= 20t4+tt(s − t)and3∂w∂t=∂w∂x∂x∂t+∂w∂y∂y∂t+∂w∂z∂z∂t=5yz +1x − ys + 25xz −1x − yt + 20xy= 80st3+s − 2tt(s − t).10. An example is the functionf(x, y) =(0 if xy = 0,1 if xy 6= 0.The partial derivatives fx(0, 0) and fy(0, 0) are both zero. However, f is not continuousat (0, 0), since approaching (0, 0) along the x-axis we have zero as the limit while ap-proaching (0, 0) along the line y = x we have 1 as the limit, so that lim(x,y)→(0,0)f(x, y)doesn’t exist.11. Define the function f(x, y) = x1/2y1/3. Then fx(x, y) =12x−1/2y1/3and fy(x, y) =13x1/2y−2/3, and in particular fx(36, 216) =12and fy(36, 216) =118. Then√353√215 ≈ f(36, 216) + dz = f (36, 216) + fx(36, 216)dx + fy(36, 216)dy= 36 +12(−1) +118(−1) = 36 −59.12. We have ∇f(x, y, z) = hy2, zey+2xy, eyi and in particular ∇f (1, 1, 1) = h1, e+2, ei. Sof is decreasing the fastest at the point (1, 1, 1) in the direction of the vector −h1, e+2, ei.13. Completing the square we havex24−(y+1)24−z24= 1 and y − 8 = z2− (x − 2)2. Thefirst is a hyperboloid of two sheets and the second is a hyperbolic paraboloid.14. Define the function f(x, y, z) = z − cos(y) sin(x). Then ∇f(x, y, z) =h−cos(y) cos(x), sin(y) sin(x), 1i and in particular ∇f(π2,π4, 0) = h0,1√2, 1i. So para-metric equations for the normal line to the surface z = c os(y) sin(x) at (π2,π4, 0) arex =π2, y =π4+1√2t, z =